/*
摘抄(简单贪心)
FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33765 Accepted Submission(s): 10999
Problem Description
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.
Input
The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.
Output
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.
Sample Input
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
Sample Output
13.333
31.500
Author
CHEN, Yue
Source
ZJCPC2004
Recommend
JGShining
简单贪心
*/
#include <iostream> //很简单的一道贪心，不知道为什么会错那么多次，搞不懂
#include <algorithm>
#include <cstdio>
#include <string.h>
#include <queue>
#include <stdlib.h>
using namespace std;
typedef struct node
{
int j;
int f;
double d;
}MC;
MC a[1003];
bool cmp(MC a,MC b)
{
return a.d>b.d;
}
int main()
{ int n,m,i;
double t;
while(scanf("%d%d",&n,&m))
{ if(n==-1&&m==-1)
break;
for(i=0;i<m;i++)
{
scanf("%d%d",&a[i].j,&a[i].f);
a[i].d=1.0*a[i].j/a[i].f;
}
sort(a,a+m,cmp);
t=0;
for(i=0;i<m;i++)
{
if(n>a[i].f)
{
t+=a[i].j;
n-=a[i].f;
}
else
{t+=n*a[i].d;break;}//老是在这出问题、郁闷---现在明白了，有当m=0时的情况，
//所以这个不能写在循环外面。
}
printf("%.3lf\n",t);
}
return 0;//现在明白了，此题的变态之处在于，m和f可以为0，伤人呀！sort居然可以排序分母为零的，强悍！！！
}
